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Thread: Sturm Liouville, orthogonality and fundamental interval

  1. #1
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    Sturm Liouville, orthogonality and fundamental interval

    Hi there. I have this exercise, which says:
    Demonstrate that:
    $\displaystyle xy''+(1-x)y'+\lambda y=0$
    has a polynomial solution for some λ values.
    Indicate the orthogonality relation between polynomials, the fundamental interval, and the weight function.


    So I thought I should solve this using Frobenius method. I have one singular point at x=0, which is regular. I assumed a solution of the form:
    $\displaystyle y(x)=\sum_0^{\infty}a_n x^{n+r}$

    And then replacing in the diff. eq. I get:
    $\displaystyle \sum_0^{\infty}a_n (n+r)(n+r-1) x^{n+r-1}+\sum_0^{\infty}a_n (n+r)x^{n+r-1}-\sum_0^{\infty}(n+r)a_n x^{n+r}+\lambda \sum_0^{\infty}a_n x^{n+r}=0$

    $\displaystyle \sum_0^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0$

    $\displaystyle a_0r^2x^{r-1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0$

    Therefore r=0.

    Then replacing r=0, and changing the index for the first summation, with m=n-1, n=m+1:
    $\displaystyle \sum_0^{\infty}a_{m+1} (m+1)^2 x^{m}-\sum_0^{\infty}a_n (n-\lambda) x^{n}=0$
    And now calling m=n
    $\displaystyle \sum_0^{\infty}x^m \left ( a_{m+1} (m+1)^2 x^{m}-a_m (m-\lambda) \right )=0$
    So I have the recurrence relation:
    $\displaystyle a_{m+1}=\frac{a_m(m-\lambda)}{(m+1)^2}$

    Trying some terms:
    $\displaystyle a_1=-a_0\lambda$

    $\displaystyle a_2=\frac{a_1(1-\lambda)}{2^2}=-\frac{a_0\lambda(1-\lambda)}{2^2}$

    $\displaystyle a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)}{2^23^2}$

    $\displaystyle a_4=\frac{a_3(3-\lambda)}{4^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)(3-\lambda)}{2^23^24^2}$


    I'm not sure what this gives, I tried this:
    $\displaystyle a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}$
    This is wrong, because the factorial in the numerator is only defined for positive values of (n-1-λ), and if n=1 I get -λ!, which wouldn't work for a_1, unless λ=0, which gives the trivial solution. But I think it works for n>1.

    So I tried in a different fashion:
    $\displaystyle a_1=-a_0\lambda$

    $\displaystyle a_2=\frac{a_1(1-\lambda)}{2^2}=\frac{a_0\lambda(\lambda-1)}{2^2}$

    $\displaystyle a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(\lambda-1)(\lambda-2)}{2^23^2}$

    $\displaystyle a_4=\frac{a_3(3-\lambda)}{4^2}=\frac{a_0\lambda(\lambda-1)(\lambda-2)(\lambda-3)}{2^23^24^2}$


    And now I called:
    $\displaystyle a_n=a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}$
    I think this is wrong too, because for example, n=1 gives $\displaystyle a_1=-a_0 \Gamma(\lambda-1)$ which doesn't fit.
    Then λ-n can't be a negative integer, and the polynomials would be given by:
    $\displaystyle \sum_0^{\infty}a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}x^n$

    Anyway, I took the diff. eq. into it's self adjoint form:

    I actually think that I didn't have to get this explicit solution. To demonstrate what the problem asks I think I should take the equation to the self adjoint form.
    $\displaystyle xy''+(1-x)y'+\lambda y=0\rightarrow y''+(\frac{1}{x}-1)y'+\frac{\lambda}{x}y=0$


    Multiplying by $\displaystyle r(x)=e^{\ln (x) -x}$
    I get:
    $\displaystyle \frac{d}{dx}\left ( e^{\ln (x) -x}\frac{dy}{dx} \right) +\lambda\frac{e^{\ln (x) -x}}{x}y=0$
    This is the self adjoint form for my differential equation. Then the weight function is given by: $\displaystyle p(x)=\frac{e^{\ln (x) -x}}{x}$

    I don't know how to get the fundamental interval.
    Would it be ok just to work it in this last form, with out finding an explicit solution?
    How can I get the fundamental interval?
    Last edited by Ulysses; Apr 2nd 2012 at 07:28 AM.
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  2. #2
    Senior Member
    Joined
    May 2010
    Posts
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    Re: Sturm Liouville, orthogonality and fundamental interval

    It's solved. I just had to use the self-adjoint form.
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